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Vilka [71]
3 years ago
12

A factory manager conducted a test on a random sample of 150 products. Of the products tested, 2 were found to be defective. Bas

ed on this information, how many products in a batch of 3000 are likely to be defective?
Mathematics
2 answers:
lions [1.4K]3 years ago
6 0

Answer:

40

Step-by-step explanation:

Write and solve a proportion:

2 / 150 = x / 3000

150x = 6000

x = 40

Phoenix [80]3 years ago
5 0

Answer:

40 products were likely to be defective.

Step-by-step explanation:

A factory worker conducted a test on a random sample of 150 products out of which 2 were found to be defected.

That means percentage of defective products was =

This concludes that 1.33% products were found to be defective.

Now based on this information we will find defective products in a batch of  3000 products.

Number of defective products = 3000×1.33% =  (3000).(1.33) over 100

Therefore 40 products were likely to be defective.

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after a party, you have 2/5 of the brownies you made left over. there are 16 left. how many brownies did you make for the party
vodomira [7]
16 = 2/5
16 ÷ 2 = 2/5 ÷ 2/1
8 = 1/5
8 * 5 = 1/5 * 5
40 = 5/5
You started the party with 40 brownies
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3 years ago
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Solve this step by step using the quadratic function <br> y=x^2 +x+2
Ne4ueva [31]

Answer:

x=<u>+</u>\frac{\sqrt{4y-7}-1 }{2}

Step-by-step explanation:

Quadratic function =( -b<u>+</u>\sqrt{b^2+4ac}  )/2a

x^2+x+2-y=0

x=-1<u>+</u>\sqrt{1^2-4(2-y)} /2

x=-1<u>+</u>\sqrt{4y-7} /2

x=<u>+</u> \frac{\sqrt{4y-7}-1 }{2}

Pretty sure this is right. Would appreciate a brainliest

3 0
2 years ago
Add the fraction 1/2+3/18
AURORKA [14]
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3/45 + 6/50 = 1/15 + 3/25 = 5/75 + 9/75 = 14/75
8 0
3 years ago
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Find the area of the shaded region. Round the nearest hundredth if necessary. YZ=14.2m
andreyandreev [35.5K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

1

   A_1  =  67.58 \ in^2

2

   A_2 =415.4 \ ft^2

3

   A_3  =  8.48 \ cm^2

4

  A_4 =  480.38 \ m^2

Step-by-step explanation:

Generally the area of a sector is mathematically represented as

         A =  \frac{\theta}{360} * \pi r^2

Now at r_1  = 11 in and  \theta_1 =  64^o

       A_1 =  \frac{64}{360} * 3.142  * 11^2

       A_1  =  67.58 \ in^2

Now at  r_2  = 20 ft in and  \theta_2  =  119 ^o

       A_2 =  \frac{119}{360} * 3.142 *  20^2

       A_2 =415.4 \ ft^2

Now at  r_3  = 6.5 cm   and  \theta_3 =  23 ^o

     A_3  =  \frac{23}{360} * 3.142 *  6.5 ^2

      A_3  =  8.48 \ cm^2

Now at  r_4  = 14.2 m   and  \theta_4 = 360 -87 =  273 ^o

         A_4 =  \frac{273}{360}  * 3.142 * 14.2^2

          A_4 =  480.38 \ m^2

6 0
3 years ago
The Perimeter is x+38
OLga [1]
P=x+38

add all individual sides together (is also the perimeter):
p=(x+3)+(x-5)+x+(x-2)+(x-4)+(x-1)+(x-1)
=7x-10

set both p equal:
x+38=7x-10
38=6x-10
48=6x
8=x

then insert x=8 in all the side/perimeter equations:
p=x+38=8+38=46

sides:
x=8
x-2=8-2=6
x-1=8-1=7
x-4=8-4=4
x-5=8-5=3
x+3=8+3=11
8 0
3 years ago
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